4.

Which of the following is a way to heat or cool a building without using electricity or another power source?

zzz [600]

I believe the answer is: A. Passive heating and cooling.

8 0

3 years ago

Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−

sergeinik [125]

Answer:

Power output, $P_{out} = 178.56 kW$

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, $T = 350^{\circ}C$

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, $\dot{m} = 0.1 kg.s^{- 1}$

Diameter of exhaust pipe, $d_{h} = 15 cm = 0.15 m$

Pressure at exhaust, P' = 50 kPa

temperature, T' = $100^{\circ}C$

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

$\dot{m} = \frac{Av_{i}}{V_{1}}$

$0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}$

$0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}$

$v_{i} = 3.986 m/s$

Now, eqn for compressible fluid:

$\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}$

Now,

$\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}$

$\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}$

$\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}$

$v_{e} = 19.33 m/s$

Now, the power output can be calculated from the energy balance eqn:

$P_{out} = -\dot{m}W_{s}$

$P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}$

$P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW$

4 0

3 years ago

Animation can occur before an action

Slav-nsk [51]

Answer:Animation can occur before an action. Prepares the audience for the action. ... A pose or action should clearly communicate to the audience the attitude, mood, reaction or idea of the character. The effective use of long, medium or close up shots, as well as camera angles helps tell the story.

Explanation:May i have brainlist plz only if u wanna give me brainlist though have an nice day and stay safe.

6 0

3 years ago

Identify the true statements about checking resistors with an ohmmeter. (Check all that apply.) Multiple select question. When c

ipn [44]

Answer:

Please find the answer in the explanation.

Explanation: To Identify the true statements about checking resistors with an ohmmeter,

When checking resistance in a circuit, there should be parallel resistance paths present. Because any circuit or resistors to be measured has to be shunted. That is connecting it in parallel so that small amounts of current can pass through the circuit.

The ohmmeter leads are connected across the resistance to be measured. An ohmmeter has its own voltage source.

The ohmmeter makes use of batteries as power source.

5 0

3 years ago

7 A very large thin plate is centered in a gap of width 0.06 m with different oils of unknown viscosities above and below; one v

sineoko [7]

Answer:

$\mu_1\approx 0.967$

$\mu_2\approx 1.933$

Explanation:

The viscosities above and below the plate are given by

$\mu_2=2\mu_1$ where $\mu_1$ and $\mu_2$ are viscosities of fluid below and above plate respectively

Force on plate due to top layer of the fluid

$\tau_1=\mu_1\frac {\triangle u}{\triangle y}$ where $\triangle u$ and $\triangle y$ are the velocity of plate and gap between the plate and upper surface respectively.

$\tau_1=\mu_1\times \frac {0.3}{0.03}=10\mu_1$

Force on plate due to bottom layer of the fluid is given by

$\tau_2=\mu_2\frac {\triangle u}{\triangle y}$ where $\triangle u$ and $\triangle y$ are the velocity of plate and gap between the plate and upper surface respectively